Question

In Laplace's rule of succession (Example 5$e$), suppose that the first$n$ flips resulted in $r$heads and$\left(n-r\right)$ tails. Show that the probability that the$\left(n+1\right)$ flip turns up heads is $\raisebox{1ex}{$\left(r+1\right)$}\!\left/ \!\raisebox{-1ex}{$\left(n+2\right)$}\right.$. To do so, you will have to prove and use the identity

${\int }_0^1 y^n(1-y{)}^{m}dy=\frac{n!m!}{(n+m+1)!}$

Hint: To prove the identity, let$C(n,m)={\int }_0^1 y^n(1-y{)}^{m}dy$ .
Integrating by parts yields

$C(n,m)=\frac{m}{n+1}C(n+1,m-1)$

Starting with$C(n,0)=1/(n+1)$ , prove the identity by induction on .

Short answer

Obtaining recursion for $C_{n,m}$ using partial integration, and then the explicit formula.

Step-by-step solution

Step: Probability wanted equation:

By choosig coin condition,

$P\left(H\mid F_{n,m}\right)=\sum _{i=0}^k P\left(H\mid C_i{F}_{n,m}\right)P\left(C_i\mid F_{n,m}\right)$

Probability head of coin,

$$\begin{gathered} P\left(H\mid C_i{F}_{n,m}\right)=P\left(H\mid C_i\right)=\frac{i}{k} \\ \begin{array}{l}P\left(C_i\mid F_{n,m}\right)=\frac{\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{i}{k}\right)}^n{\left(1-\frac{i}{k}\right)}^m}{\sum _{j=0}^k \left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{j}{k}\right)}^n{\left(1-\frac{j}{k}\right)}^m}\\ P\left(H\mid F_{n,m}\right)=\sum _{i=0}^k \frac{\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{i}{k}\right)}^{n+1}{\left(1-\frac{i}{k}\right)}^m}{\sum _{j=0}^k \left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{j}{k}\right)}^n{\left(1-\frac{j}{k}\right)}^m}\\ P\left(H\mid F_{n,m}\right)=\frac{\sum _{i=0}^k {\left(\frac{i}{k}\right)}^{n+1}{\left(1-\frac{i}{k}\right)}^m}{\sum _{j=0}^k {\left(\frac{j}{k}\right)}^n{\left(1-\frac{j}{k}\right)}^m}\end{array} \end{gathered}$$

By integral approximation of expression,

$\frac{1}{k}\sum _{i=0}^k {\left(\frac{i}{k}\right)}^n{\left(1-\frac{i}{k}\right)}^m\approx {\int }_0^1 y^n(1-y{)}^{m}dy=:C_{n,m}$

The wanted probability approximation as,

$P\left(H\mid F_{n,m}\right)\approx \frac{C_{n+1,m}}{C_{n,m}}.$

Step: 2 Partial integration:

By using partial integration,

$\begin{array}{l}{C}_{n,m}={\int }_0^1 y^n(1-y{)}^{m}dy=\left[\begin{array}{cc}{y}^{n+1}=u\left(y\right)& u^{\mathrm{\prime }}\left(y\right)=(n+1)y^n\\ (1-y{)}^m-v(y)& v^{\mathrm{\prime }}\left(y\right)=-m(1-y{)}^{m-1}\end{array}\right]\\ C_{n,m}={\int }_0^1 \frac{1}{n+1}{u}^{\mathrm{\prime }}\left(y\right)v\left(y\right)dy\\ C_{n,m}=\frac{1}{n+1}\left[{\left.\left(u\right(y\left)v\right(y\left)\right)\right|}_0^1-{\int }_0^1 u\left(y\right)v^{\mathrm{\prime }}\left(y\right)dy\right]\\ C_{n,m}=\frac{1}{n+1}\left[0-{\int }_0^1 y^{n+1}\times (-m)(1-y{)}^{m-1}dy\right]\\ C_{n,m}=\frac{m}{n+1}{\int }_0^1 y^{n+1}(1-y{)}^{m-1}dy\\ C_{n,m}=\frac{m}{n+1}{C}_{n+1,m-1}.\end{array}$

Step: 3 Proving equation:

The wanted recursion is

$\begin{array}{c}{C}_{n,m}=\frac{m}{n+1}{C}_{n+1,m-1}\\ \end{array}$

and the integration as

$C_{n,0}={\int }_0^1 y^{n}dy=\frac{1}{n+1}$

By repeating recursion as $m$times until second index reaches at $0$becomes as

$\begin{array}{l}{C}_{n,m}=\frac{m}{n+1}\times \frac{m-1}{n+2}\times \frac{m-2}{n+3}\times \dots \times \frac{1}{n+1+m-1}\times C_{n+m,0}\\ C_{n,m}=\frac{m!}{\frac{(n+m)!}{n!}\times \frac{1}{n+m+1}}\\ C_{n,m}=\frac{n!m!}{(n+m+1)!}.\end{array}$

The wanted probability as

$$\begin{gathered} \left.P\left(H\mid F_{n,m}\right)\approx \frac{C_{n+1,m}}{C_{n,m}}\right)=\frac{\frac{(n+1)!m!}{(n+m+2)!}}{\frac{n!m!}{(n+m+1)!}} \\ \left.P\left(H\mid F_{n,m}\right)\approx \frac{{\displaystyle C_{n+1,m}}}{{\displaystyle C_{n,m}}}\right)=\frac{n+1}{n+m+2}. \end{gathered}$$